Quantitative Aptitude

Geometry

1,950 Questions
All Circle Tangents and Chords 772 Coordinate Geometry 360 Circle and Arc Measures 216 Triangle Centers and Medians 175 Composite and Inscribed Shapes 174 Similarity, Congruence, and Proportionality 118 Right Triangles and Pythagorean Theorem 100 Standard 2D Mensuration 22 Polygon and Angle Sums 10 3D Solid Mensuration 3

Geometry Questions

Multiple choice

In a circle of diameter 50cm chords AB and CD are drawn parallel such that sum of their distance is 31cm and difference between their distances is 17cm.Then the length of each of the chords and A($\square$ABCD) in both the condition if AB < CD are respectively
  1. AB = 7cm, CD = 48cm, A($\square$ABCD) = 961c$m^2$ and 527 c$m^2$
  2. AB = 14cm, CD = 48cm, A($\square$ABCD) = 527c$m^2$ and 961 c$m^2$
  3. AB = 7cm, CD = 24cm, A($\square$ABCD) = 527/2c$m^2$ and 961/2 c$m^2$
  4. AB = 48cm, CD = 14cm, A($\square$ABCD) = 527c$m^2$ and 961 c$m^2$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

For a circle of radius 25 cm, a chord at distance d from the center has length 2sqrt(625 - d^2). The two distances are 24 cm and 7 cm, giving chord lengths 14 cm and 48 cm, respectively. Since AB < CD, AB = 14 cm and CD = 48 cm, with the corresponding trapezium areas 527 cm^2 and 961 cm^2.

Multiple choice

$AB$ is a diameter of a circle. $CD$ is a chord parallel to $AB$ and $2CD=AB$. The tangent at $B$ meets the line $AC$ produced at $E$ then $AE$ is equal to-
  1. $AB$
  2. $\sqrt {2} AB$
  3. $2\sqrt {2} AB$
  4. $2AB$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Using geometry of circles and similar triangles, if CD is parallel to AB and CD = 1/2 AB, the distance from the center to CD is r*sqrt(3)/2. Through coordinate geometry or similar triangles, AE evaluates to 2AB.

Multiple choice

The hypotenuse of a right triangle is $25$ cm.The difference between the lengths of the other two sides of the triangle is $5$ cm. F ind the lengths of these sides.
  1. $20$ cm and $15$ cm
  2. $12$ cm and $13$ cm
  3. $21$ cm and $12$ cm
  4. $13$ cm and $18$ cm
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Let the sides be x and x-5. By the Pythagorean theorem, x^2 + (x-5)^2 = 25^2. Solving x^2 + x^2 - 10x + 25 = 625 gives 2x^2 - 10x - 600 = 0, or x^2 - 5x - 300 = 0. Factoring gives (x-20)(x+15) = 0, so x=20 and the other side is 15.

Multiple choice

The length of the hypotenuse of a right triangle exceeds the length of the base by $2$ cm and exceeds twice the length of the attitude by $1$ cm. The perimeter of the triangle is:
  1. $18$ cm
  2. $17$ cm
  3. $25$ cm
  4. $40$ cm
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Let the altitude be h, the base be b, and the hypotenuse be c. The conditions give c = b + 2 and c = 2h + 1, so b = 2h - 1; applying the Pythagorean theorem gives h = 8, b = 15, and c = 17. The perimeter is 8 + 15 + 17 = 40 cm.

Multiple choice

The length of a hypotenuse of a right triangle exceeds the length of its base by $2 cm.$ and exceeds twice the length of the altitude by $1 cm$. Find the length of each side of the triangle (in $cm$).
  1. $6, 8, 10$
  2. $7, 24, 25$
  3. $8, 15, 17$
  4. $7, 40, 41$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Let base = b, altitude = a, hypotenuse = h. h = b + 2. h = 2a + 1. So b = 2a - 1. By Pythagoras: (2a-1)^2 + a^2 = (2a+1)^2. 4a^2 - 4a + 1 + a^2 = 4a^2 + 4a + 1. a^2 - 8a = 0. a = 8. Then b = 15, h = 17.

Multiple choice

Calculate the length of the hypotenuse of a right-angled triangle if one leg is $3$ inches longer than the other one and the area of the triangle is $35$ square inches.
  1. $10$
  2. $\displaystyle\frac{\sqrt{35}}{2}$
  3. $7\sqrt{10}$
  4. $\sqrt{149}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Let the legs be x and x+3. Area = 0.5 * x * (x+3) = 35, so x^2 + 3x - 70 = 0. Solving gives x = 7 or -10. The legs are 7 and 10. The hypotenuse is sqrt(7^2 + 10^2) = sqrt(49 + 100) = sqrt(149).

Multiple choice

A circle is inscribed in an equilateral triangle with side lengths $6$ unit. Another circle is drawn inside the triangle (but outside the first circle), tangent to the first circle and two of the sides of the triangle, the radius of the smaller circle is?
  1. $1/\sqrt{3}$
  2. $2/3$
  3. $1/2$
  4. $1$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The inradius of an equilateral triangle with side 6 is r = side / (2 * sqrt(3)) = 6 / (2 * sqrt(3)) = sqrt(3). The smaller circle is tangent to the larger circle and two sides. Using the property of circles in an equilateral triangle, the radius of the smaller circle is 1/3 of the larger one. r_small = sqrt(3) / 3 = 1/sqrt(3).

Multiple choice

If in a $\Delta ABC$, $b=12$ units, $c=5$ units and $\Delta = 30$ sq. units, then the distance between vertex $A$ and incentre of the triangle is equal to?
  1. $2$ units
  2. $2\sqrt2 $ units
  3. $\sqrt2 $ units

  4. none of these
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Incenter distance from vertex A is given by r / sin(A/2). Area = 30, s = (a+b+c)/2. Need a. Area = 0.5 * b * c * sin(A) => 30 = 0.5 * 12 * 5 * sin(A) => sin(A) = 1. So A = 90 degrees. a = sqrt(12^2 + 5^2) = 13. s = (13+12+5)/2 = 15. r = Area/s = 30/15 = 2. Distance = 2 / sin(45) = 2 / (1/sqrt(2)) = 2 * sqrt(2).

Multiple choice

If $O$ is circumcentre of $\triangle ABC$ and ${R}{1},{R}{2},{R}{3}$ are circumradii of triangles $OBC,OCA$ and $OAB$, then $\displaystyle \frac { a }{ { R }{ 1 } } +\frac { b }{ { R }{ 2 } } +\frac { c }{ { R }{ 3 } } =$
  1. $\displaystyle \frac { abc }{ { R }^{ 3 } } $
  2. $\displaystyle \frac { 2abc }{ { R }^{ 3 } } $
  3. $\displaystyle \frac { abc }{ { 2R }^{ 3 } } $

  4. none of these
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

In a triangle, a/sinA = 2R. For OBC, the circumradius R1 = a / (2*sin(angle BOC)). Angle BOC = 2A. So R1 = a / (2*sin(2A)). This leads to a/R1 = 2*sin(2A) = 4*sinA*cosA. Summing these is a known identity related to the circumcenter, resulting in abc/R^3.